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QM inverse problem

  1. Jul 11, 2013 #1

    hilbert2

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    Let's say we have a one dimensional single particle system that is described by a SE

    -[itex]\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi(x)+V(x)\psi(x)=E\psi(x)[/itex]

    We do not know what the potential energy function V(x) is, but we know that the eigenvalues
    spectrum is

    [itex]E_{n}=kn^{\frac{3}{2}}[/itex]

    where k is a constant with dimensions of energy and n=1,2,3,...

    How to find out what is the function V(x) that results in this energy spectrum? If the exponent
    were 1 instead of 3/2 then it would be a harmonic oscillator potential and if the exponent were
    2 it would be the "particle in box" potential. Do all possible exponents result in a V(x) that can be
    written with elementary functions?

    To find V(x), we should obviously

    1. Write the Hamiltonian matrix in the basis where its diagonal (we can do that)

    2. Unitary transform to position representation basis

    3. Subtract the kinetic energy operator

    But how do we find the right unitary transformation to do this?
     
    Last edited: Jul 11, 2013
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  3. Jul 11, 2013 #2

    mfb

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    I would test V(x)=|x| and if that does not work, the more general V(x)=|x|a with real a, but that is just a guess.
    In the general case: No idea. I know that in scattering theory, a transformation of the spatial potential to a "momentum potential" helps, but I don't know if that can be used here.
     
  4. Jul 11, 2013 #3

    hilbert2

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    ^ I don't right now have at hand a numerical code that solves for the eigenstates and eigenvalues in 1D potential, but if I have time to make one I'll try what happens with the V(x) = |x| potential.

    One has to note that the energy spectrum does not uniquely determine V(x). For example, the energy levels are same in the harmonic potential V(x) = kx2 and the translated harmonic potential V(x) = k(x+c)2 (c is a constant). Also applying a parity transformation to V(x) does not change the energy levels.

    Also, its obvious that some conceivable spectra don't correspond to any physical potential. Suppose the energy spectrum is bounded from upward and En converges to some maximum energy when n grows without bound. That is not physically possible.
     
  5. Jul 11, 2013 #4

    hilbert2

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    I did some searching with Google and found a lot of mathematical literature about this exact problem (reconstructing a potential from spectral or scattering data). Look like the mathematicians Gelfand, Levitan and Marchenko solved this problem in 1960's.

    I didn't understand much of the literature as it was for mathematicians, and it didn't provide any simple recipe for finding V(x) when En is known...
     
  6. Jul 11, 2013 #5

    mfb

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    That's exactly how the energy states for a 1/r2 potential (in atoms, for example) look like.
     
  7. Jul 11, 2013 #6

    hilbert2

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    That's only the bound states. The scattering states where the electron's energy is higher than the ionization energy of hydrogen (for example), don't have an energy upper limit. Scattering states form the continuum part of the hydrogenic energy spectrum.
     
  8. Jul 11, 2013 #7

    mfb

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    Sure, but if you include unbound states you never get an upper limit on the energy.

    I would not call this "conceivable spectra" ;).
    The same goes for spectra without a lower bound.
     
  9. Jul 11, 2013 #8

    hilbert2

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    I vaguely remember from a QFT course that the fact that Dirac field operators must anticommute, is proven by arguing that otherwise the field modes could have arbitrarily large negative energy...

    Of course if the energy levels of some system went towards negative infinity, you could extract arbitrarily large amounts of usable energy from the system by forcing it into lower and lower energy levels.
     
  10. Jul 11, 2013 #9

    Avodyne

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    You can get an approximate answer via the semiclassical Weyl formula for the number of states N(E) below a given energy E:
    [tex]N(E) = \int{dx \,dp\over 2\pi\hbar}\,\theta(E-H(x,p))[/tex]
    where [itex]\theta(z)[/itex] is the unit step function, [itex]H(x,p)[/itex] is the classical hamiltonian, and [itex]x[/itex] and [itex]p[/itex] are both integrated from -∞ to +∞.

    I suspect this is what is intended. An exact answer would be much too hard.
     
  11. Jul 12, 2013 #10

    hilbert2

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    In the problem, I already know what the function N(E) is, and I want to find the classical hamiltonian. I would have to somehow invert that formula.
     
  12. Jul 12, 2013 #11

    Avodyne

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    Try V(x) = g|x|^n, and find the value of n that gives the N(E) that you want.
     
  13. Jul 13, 2013 #12
    Actually it can be obtained in an approximate way by wkb quantization condition.If you choose a potential v(x)=∞ for x<0 and v(x)=mgx for x>0 and solve it using wkb quantisation condition(by taking into account the infinite discontinuity otherwise it will get wrong).This potential can be solved exactly using Airy functions but nonetheless an approximate solution(using wkb) will give a dependence like [n+(1/4)]3/2 for the eigenvalue.
     
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